4 cards are drawn at random from a standard 52-card deck. At least 3 of those cards are spades. What is the probability that all 4 cards are spades?
I know that (one possible) correct solution to the problem is $\frac{\binom{13}{4}}{\binom{13}{4} + 39\binom{13}{3}} = 5/83$. However, my interest instead lies in seeing where I overcount if I instead do $\frac{\binom{13}{4}}{49\binom{13}{3}} = \frac{5}{98}$.
The logic for the correct solution is simple: (first four cards are 4 spades in any order) divided by the quantity (first four cards are 4 spades in any order) + (first 3 cards are spades in any order and the fourth card is not a spade).
The logic for the erroneous solution is similar. There are $\binom{13}{3}$ ways to choose 3 cards that are definitely spades. There remain $52-3=49$ cards that are anything, including spades. Therefore, we should have "at least" 3 spades, right?
Where does that logic go wrong?
If four cards are drawn from a standard deck and at least three cards are spades, then there are two possibilities: three spades and one non-spade, which can be selected in $\binom{13}{3}\binom{39}{1}$ ways (without regard to the order of selection), or four spades, which can be selected in $\binom{13}{4}$ ways. Hence, the conditional probability that four spades are selected given that at least spades are selected when four cards are drawn from a standard $52$-card deck is indeed $$\frac{\dbinom{13}{4}}{\dbinom{13}{3}\dbinom{39}{1} + \dbinom{13}{4}}$$
Your second method is incorrect since you are counting each hand with four spades four times, once for each way you could designate three of the spades in your hand as the three spades in the hand. For instance, suppose your hand consists of $5\spadesuit, 7\spadesuit, 10\spadesuit, Q\spadesuit$. You count this hand four times:
$$\begin{array}{c c} \text{three spades} & \text{extra card} \\ \hline 5\spadesuit, 7\spadesuit, 10\spadesuit & Q\spadesuit\\ 5\spadesuit, 7\spadesuit, Q\spadesuit & 10\spadesuit\\ 5\spadesuit, 10\spadesuit, Q\spadesuit & 7\spadesuit\\ 7\spadesuit, 10\spadesuit, Q\spadesuit & 5\spadesuit \end{array} $$
Notice that $$\binom{13}{3}\binom{39}{1} + \color{red}{4}\binom{13}{4} = \color{red}{49}\binom{13}{3}$$